Solve 5x^2-9x-12=

$$5 { x }^{ 2 } -9x-12=$$

Factor

$$5\left(x-\frac{9-\sqrt{321}}{10}\right)\left(x-\frac{\sqrt{321}+9}{10}\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$5x^{2}-9x-12=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 5\left(-12\right)}}{2\times 5}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 5\left(-12\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-9\right)±\sqrt{81-20\left(-12\right)}}{2\times 5}$$

Multiply -20 times -12.

$$x=\frac{-\left(-9\right)±\sqrt{81+240}}{2\times 5}$$

Add 81 to 240.

$$x=\frac{-\left(-9\right)±\sqrt{321}}{2\times 5}$$

The opposite of -9 is 9.

$$x=\frac{9±\sqrt{321}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{9±\sqrt{321}}{10}$$

Now solve the equation x=\frac{9±\sqrt{321}}{10} when ± is plus. Add 9 to \sqrt{321}.

$$x=\frac{\sqrt{321}+9}{10}$$

Now solve the equation x=\frac{9±\sqrt{321}}{10} when ± is minus. Subtract \sqrt{321} from 9.

$$x=\frac{9-\sqrt{321}}{10}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{9+\sqrt{321}}{10} for x_{1} and \frac{9-\sqrt{321}}{10} for x_{2}.

$$5x^{2}-9x-12=5\left(x-\frac{\sqrt{321}+9}{10}\right)\left(x-\frac{9-\sqrt{321}}{10}\right)$$

Evaluate

$$5x^{2}-9x-12$$